# Potential function for conservative internal forces

In Goldstein's Classical Mechanics book, he considered a system of particles and looked at the conservative internal force between particle $$i$$ and particle $$j$$ that satisfy the strong law of action and reaction.

He wrote the potential function for this internal force as $$V_{ij}=V_{ij}(|\vec{r_i}-\vec{r_j}|) .$$

He then said that the forces $$\vec{F}_{ji}$$ (force particle j exerts on i) and $$\vec{F}_{ij}$$ (force particle i exerts on j) are automatically equal and opposite:

$$\vec{F}_{ji} = - \nabla_i V_{ij}(|\vec{r_i}-\vec{r_j}|) = + \nabla_j V_{ij}(|\vec{r_i}-\vec{r_j}|) =-\vec{F}_{ij} .$$

I have some problems seeing why $$- \nabla_i V_{ij}(|\vec{r_i}-\vec{r_j}|) = + \nabla_j V_{ij}(|\vec{r_i}-\vec{r_j}|).$$The gradient operator acts on different indices, why does a change of sign makes them equal?

\begin{align} \dfrac{\partial}{\partial \vec{r}_i}V_{ij}(\vert \vec{r}_i-\vec{r}_j\vert)&=\dfrac{\partial V_{ij}}{\partial (\vec{r}_i-\vec{r}_j)}\dfrac{\partial (\vec{r}_i-\vec{r}_j)}{\partial \vec{r}_i}\\ &=\dfrac{\partial V_{ij}}{\partial (\vec{r}_i-\vec{r}_j)}(1-\delta_{ji}) \end{align}

\begin{align} \dfrac{\partial}{\partial \vec{r}_j}V_{ij}(\vert \vec{r}_i-\vec{r}_j\vert)&=\dfrac{\partial V_{ij}}{\partial (\vec{r}_i-\vec{r}_j)}\dfrac{\partial (\vec{r}_i-\vec{r}_j)}{\partial \vec{r}_j}\\&=\dfrac{\partial V_{ij}}{\partial (\vec{r}_i-\vec{r}_j)}(\delta_{ij}-1)\\&=-\dfrac{\partial}{\partial \vec{r}_i}V_{ij}(\vert \vec{r}_i-\vec{r}_j\vert) \end{align}
As far as $$\delta_{ij}=\delta_{ji}.$$
• Can you explain why you replaced $\nabla_i$ with $\partial \over {\partial \vec{r}_i}$? My understanding is that $\nabla_i = \frac{\partial}{\partial x_i} \vec{i } + \frac{\partial}{\partial y_i} \vec{j } + \frac{\partial}{\partial z_i} \vec{k }$ May 22, 2020 at 15:33